Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $q \neq 0$. $x = \dfrac{q(4q + 3)}{10} \times \dfrac{8}{20q^2 + 15q} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $x = \dfrac{ q(4q + 3) \times 8 } { 10 \times (20q^2 + 15q) } $ $ x = \dfrac {8 \times q(4q + 3)} {10 \times 5q(4q + 3)} $ $ x = \dfrac{8q(4q + 3)}{50q(4q + 3)} $ We can cancel the $4q + 3$ so long as $4q + 3 \neq 0$ Therefore $q \neq -\dfrac{3}{4}$ $x = \dfrac{8q \cancel{(4q + 3})}{50q \cancel{(4q + 3)}} = \dfrac{8q}{50q} = \dfrac{4}{25} $